Page 177 - Shigley's Mechanical Engineering Design

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152 Mechanical Engineering Design
loading are symmetric relative to the midspan. However, we will use the given bound-
ary conditions of the problem and verify that the slope is zero at the midspan. Integrating
Eq. (1) gives
wl 3 w 4

EIy = Mdx = x − x + C 1 x + C 2 (2)
12 24
The boundary conditions for the simply supported beam are y = 0 at x = 0 and l.
Applying the first condition, y = 0 at x = 0, to Eq. (2) results in C 2 = 0. Applying the
second condition to Eq. (2) with C 2 = 0,
wl 3 w 4
EIy(l) = l − l + C 1 l = 0
12 24
3
Solving for C 1 yields C 1 =−wl /24. Substituting the constants back into Eqs. (1) and
(2) and solving for the deflection and slope results in
wx 2 3 3
y = (2lx − x − l ) (3)
24EI
dy w 2 3 3
θ = = (6lx − 4x − l ) (4)
dx 24EI
Comparing Eq. (3) with that given in TableA–9, beam 7, we see complete agreement.
For the slope at the left end, substituting x = 0 into Eq. (4) yields

wl 3
θ| x=0 =−
24EI
and at x = l,
wl 3
θ| x=l =
24EI
At the midspan, substituting x = l/2 gives dy/dx = 0, as earlier suspected.
The maximum deflection occurs where dy/dx = 0. Substituting x = l/2 into
Eq. (3) yields
5wl 4
y max =−
384EI
which again agrees with Table A–9–7.

The approach used in the example is fine for simple beams with continuous
loading. However, for beams with discontinuous loading and/or geometry such as a step
shaft with multiple gears, flywheels, pulleys, etc., the approach becomes unwieldy. The
following section discusses bending deflections in general and the techniques that are
provided in this chapter.
4–4 Beam Deflection Methods

Equations (4–10) through (4–14) are the basis for relating the intensity of loading q,
vertical shear V, bending moment M, slope of the neutral surface θ, and the trans-
verse deflection y. Beams have intensities of loading that range from q = constant